3.185 \(\int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=35 \[ \frac {(a+b x) \log (a+b x)}{b \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

(b*x+a)*ln(b*x+a)/b/((b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {608, 31} \[ \frac {(a+b x) \log (a+b x)}{b \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((a + b*x)*Log[a + b*x])/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 608

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {1}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(a+b x) \log (a+b x)}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.74 \[ \frac {(a+b x) \log (a+b x)}{b \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((a + b*x)*Log[a + b*x])/(b*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.96, size = 10, normalized size = 0.29 \[ \frac {\log \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

log(b*x + a)/b

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giac [A]  time = 0.15, size = 17, normalized size = 0.49 \[ \frac {\log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

log(abs(b*x + a))*sgn(b*x + a)/b

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maple [A]  time = 0.04, size = 25, normalized size = 0.71 \[ \frac {\left (b x +a \right ) \ln \left (b x +a \right )}{\sqrt {\left (b x +a \right )^{2}}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x+a)^2)^(1/2),x)

[Out]

(b*x+a)*ln(b*x+a)/b/((b*x+a)^2)^(1/2)

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maxima [A]  time = 1.33, size = 12, normalized size = 0.34 \[ \frac {\log \left (x + \frac {a}{b}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

log(x + a/b)/b

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mupad [B]  time = 0.21, size = 19, normalized size = 0.54 \[ \frac {\ln \left (a+b\,x+\sqrt {{\left (a+b\,x\right )}^2}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^2)^(1/2),x)

[Out]

log(a + b*x + ((a + b*x)^2)^(1/2))/b

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sympy [A]  time = 0.09, size = 7, normalized size = 0.20 \[ \frac {\log {\left (a + b x \right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)**2)**(1/2),x)

[Out]

log(a + b*x)/b

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